Let \(\vec a\) and \(\vec b\) be two unit vectors such that \(|\vec a - \vec b|<2.\) If 2θ is the angle between \(\vec a\) and \(\vec b,\) then which one of the following is correct ?

Concept:

1. The scalar product of two vectors a and b of magnitude |a| and |b| is given as |a||b| cos θ, where θ represents the angle between the vectors a and b taken in the direction of the vectors.
2. We can express the scalar product as \(\mid\vec a\mid.\mid\vec b\mid\)=|a||b| cosθ, where |a| and |b| represent the magnitude of the vectors a and b while cos θ denotes the cosine of the angle between both the vectors and a.b indicates the dot product of the two vectors.

Calculation:

Given:

\(\vec a\) and \(\vec b\) be two unit vectors such that \(|\vec a - \vec b|<2\) and 2θ is the angle between \(\vec a\) and \(\vec b\)

⇒ \(\mid\vec a\mid= \mid\vec b\mid=1\)

We are given \(|\vec a - \vec b|<2.\)

Squaring both sides, we get, 

⇒ \(|\vec a - \vec b|^2<2^2\)

⇒ \((\vec a - \vec b)(\vec a - \vec b)<4\)

⇒ \(\mid\vec a\mid^2+ \mid\vec b\mid^2-2\ \vec a.\vec b<4\)

⇒ 1 + 1 - 2 \(\mid\vec a\mid. \mid\vec b\mid\ cos2θ<4\)

⇒ 1 + 1 - 2​ . 1 . 1 . cos2θ < 4

⇒ 2 - 2.cos2θ < 4

⇒ 1 - cos2θ < 2

⇒ 2 sin²θ < 2

⇒ sin2θ < 1

∴ -1≤ sin θ < 1 is correct.